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Can someone help me in this trig question, please? thanks
A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12

Can someone help me in this trig question please thanks A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise a class=
Can someone help me in this trig question please thanks A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise a class=

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[tex]\bf \textit{the position of the rider is clearly }20cos\left( \frac{5\pi }{12} \right)~~,~~20sin\left( \frac{5\pi }{12} \right)\\\\ -------------------------------\\\\ \cfrac{5}{12}\implies \cfrac{2+3}{12}\implies \cfrac{2}{12}+\cfrac{3}{12}\implies \cfrac{1}{6}+\cfrac{1}{4} \\\\\\ \textit{therefore then }\qquad \cfrac{5\pi }{12}\implies \cfrac{1\pi }{6}+\cfrac{1\pi }{4}\implies \cfrac{\pi }{6}+\cfrac{\pi }{4}\\\\ -------------------------------[/tex]

[tex]\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\ -------------------------------\\\\ cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=cos\left( \frac{\pi }{6}\right)cos\left(\frac{\pi }{4} \right)-sin\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)[/tex]

[tex]\bf cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}-\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{6}}{4}-\cfrac{\sqrt{2}}{4}\implies \boxed{\cfrac{\sqrt{6}-\sqrt{2}}{4}} \\\\\\ sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=sin\left( \frac{\pi }{6}\right)cos\left( \frac{\pi }{4} \right)+cos\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)[/tex]

[tex]\bf sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}+\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{2}}{4}+\cfrac{\sqrt{6}}{4}\implies \boxed{\cfrac{\sqrt{2}+\sqrt{6}}{4}}\\\\ -------------------------------\\\\ 20\left( \cfrac{\sqrt{6}-\sqrt{2}}{4} \right)\implies 5(-\sqrt{2}+\sqrt{6}) \\\\\\ 20\left( \cfrac{\sqrt{2}+\sqrt{6}}{4} \right)\implies 5(\sqrt{2}+\sqrt{6})[/tex]
oladayoademosu

The exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

The position

Since the position of the carousel is (x, y) = (20cosθ, 20sinθ) and we need to find the position when θ = 5π/12 = 5π/12 × 180 = 75°

So, substituting the value of θ into the positions, we have

(20cos75°, 20sin75°)

The value of 20cos75°

20cos75° = 20cos(45 + 30)

Using the compound angle formula

cos(A + B) = cosAcosB - sinAsinB

With A = 45 and B = 30

cos(45 + 30) = cos45cos30 - sin45sin30

= 1/√2 × √3/2 - 1/√2 × 1/2

= 1/2√2(√3 - 1)

= 1/2√2(√3 - 1) × √2/√2

= √2(√3 - 1)/4

= (√6 - √2)/4

= (-√2 + √6)/4

So, 20cos75° = 20 × (-√2 + √6)/4

= 5 (-√2 + √6)

The value of 20sin75°

20sin75° = sin(45 + 30)

Using the compound angle formula

sin(A + B) = sinAcosB + cosAsinB

With A = 45 and B = 30

sin(45 + 30) = sin45cos30 + cos45sin30

= 1/√2 × √3/2 + 1/√2 × 1/2

= 1/2√2(√3 + 1)

= 1/2√2(√3 + 1) × √2/√2

= √2(√3 + 1)/4

= (√6 + √2)/4

= (√2 + √6)/4

So, 20sin75° = 20 × (√2 + √6)/4

= 5(√2 + √6)

Thus, (20cos75°, 20sin75°) = 5 (-√2 + √6), 5(√2 + √6).

So, the exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

Learn more about position here:

https://brainly.com/question/11001232

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